Unit 6 Chemical Analysis

Unit 6 Chemical Analysis Chapter 8

Objectives 39Perform calculations using the mole to calculate the molar mass 40Perform calculations using the mole to convert between grams, number of particles, volume, and moles 41Perform advanced calculations using empirical formula, molecular formula, and percent composition

The Mole • Society prefers to work with simple numbers. • When a very large or very small numbers are involved, society converts them to easier numbers. • For example: • 2 shoes = 1 pair • 12 eggs = 1 dozen • 10 years = 1 decade • 100 years = 1 century

The Mole • Science operates under the same idea. • As that atoms are so small, their average atomic mass is very hard to work with in lab. • A scientist by the name of Amedeo Avogadro proposed an idea that a given volume of gas is proportional to the number of atoms or molecules regardless of the chemical identity of the gas assuming a constant temperature and pressure were held.

The Mole • Avogadro’s idea inspired scientists to look for the exact number of atoms or molecules in a given volume. • In the mid-1800s, scientists first determined the number. • Over time, the number was modified slightly to be more accurate and is now: 6.022141 x 1023

The Mole • The number was called Avogadro’s constant. • Because the number was so large, the number was set equal to 1 mole and thus we can say: 1 mole = 6.02 x 1023 particles Often Avogadro’s constant is written as 6.02 x 1023 instead of writing out all of the known numbers.

The Mole • One of the more useful aspects of the mole is what it can do for masses. • Since Avogrado’s number is a proportionality factor molar mass to actual mass, it allows for conversions from amu’s to grams. • For instance, • It is known that a proton has a mass of 1 atomic mass unit. • If there were 6.02 x 1023 protons, it would have a mass of 1 gram.

Calculating molar mass • The molar mass of a chemical looks at how much mass it would have if there was 6.02 x 1023 units of that chemical. • For instance, • Assume, we have 1 atom of sodium. • Sodium has a mass of 22.99 amu’s • If there were 6.02 x 1023 sodium atoms (or 1 mole) it’s mass would be 22.99 grams. • We would say that sodium has a molar mass of 22.99 grams/mole.

Calculating molar mass • The molar mass of a compound or molecule works the same way. • Assume, we had carbon monoxide. • If we had one molecule, there would be 1 carbon atom and 1 oxygen atom. • To deteminethe mass of the molecule, we would add the mass of each atom (12.01 amu + 16.00 amu = 28.01 amu’s) • Now if we had 6.02 x 1023 molecules of carbon monoxide, we would have a mass of 28.01 grams. • Thus the molar mass of carbon monoxide is 28.01 grams/mole. Practice

Molar Masses Calculate the molar masses of the following: • Mg • H2O • CaCl2 • Mg(NO3)2

Molar Masses Calculate the molar masses of the following: • Mg 24.31 grams/mole (1 Mg x 24.31g) • H2O 18.02 grams/mole (2 H x 1.01 g) + (1 O x 16.00 g) • CaCl2 110.98 grams/ mole (1 Ca x 40.08 g) + (2 Cl x 35.45 g) • Mg(NO3)2 148.33 grams/mole (1 Mg x 24.31g) + (2 N x 14.01g) + (6 O x 16g) Return

Using the mole • In science, it is rare to find exactly one mole of a substance or to have the exact molar mass. • Therefore, it is necessary to be able to convert from molar mass to moles or to particles or to volume depending on what you are looking for. • In order to do this, we use dimensional analysis.

Dimensional Analysis • In Unit 1, we learned a technique for converting known as dimensional analysis. • It had the basic setup shown below: • The conversion factor to remember is: 1 mole = 6.02 x 1023 particles = Molar Mass (g) = 22.4 L *Particles can stand for atoms, molecules, or formula units. Conversion Factor

Using the mole • To see how dimensional analysis works, assume we have 15 grams of water, and we would want to know how many moles this is. • We know the given is 15 grams of water • We can calculate the water has a molar mass of 18.02 grams/mole (our conversion factor) • Since we want moles, that unit will go on top. • We have grams so that unit will go on the bottom. • At this point, we can calculate. 15 g water 1 mole water 0.83 moles water 18.02 g water

Using the mole • As long as you remember the conversion factor and the setup for dimensional analysis, you should be able to convert. • Click the link to the left for additional practice. Practice

Mole Conversions • Convert 2.5 moles to atoms: • Convert 24 grams CO2 to moles: • Convert 6.75 x 1022 molecules to moles: • Convert 48 liters to grams:

Mole Conversions • Convert 2.5 moles to atoms: • Convert 24 grams CO2 to moles: • Convert 6.75 x 1022 molecules to moles: • Convert 48 liters to grams of hydrogen gas: Return 4.3 g H2

Chemical Analysis • A large part of chemistry is analyzing unknown materials. • There are several analytical instruments that help with this process that will be talked about later in the year. • We are going to take a look at how to use the data these instruments provide.

Percent Composition • In Unit 4, we discussed the relative size of atoms. • When a compound is created, such as KF, the size can help us visualize what percent is potassium and percent is fluorine. • In this case, it is clear that even though we have one of each atom, potassium makes up a larger percentage of this compound. K+ F-

Percent Composition • A more exact percentage can be calculated using the molar mass of the compound. • KF would have a molar mass of 58.10 g/mol. • If we divide the mass of each element by the whole, we will get the percentage of each:

Percent Composition-Review • Calculate the molar mass of the compound. • Calculate the total mass of each element in the compound. • Divide the total mass of each element by the molar mass of the compound. • Multiply that answer by 100 Practice

Percent Composition • Calculate the % composition of CaCl2: • Calculate the % composition of CaSO4:

Percent Composition • Calculate the % composition of CaCl2: • Calculate the % composition of CaSO4: Return

Empirical Formula • When doing a chemical analysis, the percent composition is often what is provided. • The empirical formula is the lowest whole number ratio of the elements that make up the compound.

Empirical Formula • Assume we are given the following analysis for a chemical: • The chemical is 13.20% magnesium. • The chemical is 86.80% bromine. • We want to know the empirical formula. • Once we have the %’s, assume you have exactly 100. grams for the analysis. • By doing this, we can easily convert the %’s into grams. • If I have 13.20% of 100. grams, I have 13.20 grams.

Empirical Formulas • Since we assumed we had 100. grams, • We can say that we have 13.20g of Mg and 86.80g of Br. • Our goal is to find a ratio of Mg to Br. Since this is looking for a number of each element, we can use the mole to determine this ratio.

Empirical Formulas • Now we know how many moles of each we have: • 0.5430 moles Mg • 1.086 moles Br • We can now look at the ratio between the two. • Since we want a whole number ratio, we need to modify our numbers. • To do this, divide each by the smaller number. • 0.5430/0.5430 = 1Mg • 1.086/0.5430 = 2 Br Therefore, our empirical formula would be MgBr2.

Empirical Formulas-Review • Convert the %’s to grams. • Assume 100 grams • Convert grams into moles for each element. • Divide each part by the smallest number from step #2. • Round to nearest whole number if close • i.e.: 3.99 should be written as 4 • The answers from step #3 are the subscripts for each element. Practice

Determine the empirical formulas • 11.2% H and 88.8% O • 36.48% Na, 25.44% S, and 38.08% O

Determine the empirical formulas • 11.2% H and 88.8% O • 11.2 g H and 88.8 g O • 11.1 / 5.55 = 2 H 5.55 / 5.55 = 1 O H2O

Determine the empirical formulas • 36.48% Na, 25.44% S, and 38.08% O • 36.48 g Na, 25.44 g S and 38.08 g O • Na • 1.587 / 0.7933 = 2 Na 0.7933 / 0.7933 = 1 S 2.389 / 0.7933 = 3 O Na2SO3 Return

Molecular Formulas • Calculating the empirical formula will always work if your compound is ionic. • However, there are a few more steps to determining covalent or organic formulas from the percent composition. • Here is the problem: • If there was 14.4% H and 85.6% C, the empirical formula would be CH2. • Unfortunately, C2H4, C3H6, C4H8,…. will all give a percent composition of 14.4% H and 85.6 % C and they are all known molecules.

Molecular Formulas • One other piece of information that is obtained while doing a chemical analysis is the typically the mass of the unknown compound. • If we know the mass of the compound and the empirical formula, we will be able to determine the correct molecular formula.

Molecular Formulas • For the example from before, we knew that the empirical formula was CH2. • Assume, our analysis also provided the mass of the unknown compound to be 42.09 grams. • If we divide the mass of the unknown by the molar mass of the empirical formula we will determine how many CH2’s make up this molecule. Therefore, the compound is 3(CH2). Since we do not write formulas in that way, multiple the 3 by each subscript to get C3H6.

Molecular Formula-Review • Determine the empirical formula. • Divide the mass of the unknown by the molar mass of the empirical formula. • Multiple the subscripts on the empirical by the answer to step #2. Practice

Determine the molecular formula • Empirical formula CH3 • Mass of unknown compound: 60.16 grams • Empirical Formula SiH3Cl • Mass of unknown compound: 199.70 grams

Determine the molecular formula • Empirical formula CH3 • Mass of unknown compound: 60.16 grams C4H12 • Empirical Formula SiH3Cl • Mass of unknown compound: 199.70 grams Si3H9Cl3 Return

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Unit 6 Chemical Analysis